Locus and the Parabola question (1 Viewer)

[Aron]

Find the equation of the parabola with vertex (-2,3) that also passes through (2,1) and is concave downwards.

Thanks in advance

 

[braintic]

(x+2)^2 = 4a(y-3)

Substitute (2,1):

16 = -8a
a=-2

So equation is (x+2)^2 = -8(y-3)

 

[QZP]

How can focal LENGTH be negative?? :S

 

[hamstar]

How can focal LENGTH be negative?? :S

well in the above solution, it is supposed to be (x+2)^2=-4a(y-3)

(this is because it is concave down)

and then u sub in the points (2,1) and you get

16=8a

hence a=2

therefore the equation is (x+2)^2 = -8(y-3)

 

[QZP]

I know Just wanted people to notice the fallacy.

 

[braintic]

I know Just wanted people to notice the fallacy.

There is no requirement for 'a' to represent the focal length unless it is defined that way in the question.

 

[Aron]

thanks, now i see to sub in the vertex co-ordinates in the equation of the parabola. Cheers for the reply